Integral of a differential form

  • #1
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Homework Statement



Suppose that a smooth differential ##n-1##-form ##\omega## on ##\mathbb{R}^n## is ##0## outside of a ball of radius ##R##. Show that $$
\int_{\mathbb{R}^n} d\omega = 0.
$$

Homework Equations


[/B]
$$\oint_{\partial K} \omega = \int_K d\omega$$

The Attempt at a Solution



If ##\omega## is ##0## outside of the ball, by continuity, it must be ##0## on the surface of the ball as well. We know that $$
\omega = \sum_{i=1}^n f^i(x^1,\ldots,x^n) dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n
$$ for some functions ##f^i:\mathbb{R}^n\to\mathbb{R}##, so
$$
d\omega = \sum_{i=1}^n (-1)^{i-1} \frac{\partial f^i}{\partial x^i} dx^1\wedge\cdots\wedge dx^n.
$$
All those partial derivatives are ##0## outside of the ball, so $$
\int_{\mathbb{R}^n} d\omega = \int_B d\omega = \oint_{\partial B} \omega = 0,
$$ where ##B## is the ball.
 

Answers and Replies

  • #2
Orodruin
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